/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A baseball coach uses a radar de... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 37: Problem 68

A baseball coach uses a radar device to measure the speed of an approaching pitched baseball. This device sends out electromagnetic waves with frequency \(f_{0}\) and then measures the shift in frequency \(\Delta f\) of the waves reflected from the moving baseball. If the fractional frequency shift produced by a baseball is \(\Delta f / f_{0}=2.86 \times 10^{-7}\) , what is the baseball's speed in \(\mathrm{km} / \mathrm{h} ?\) (Hint: Are the waves Doppler- shifted a second time when reflected off the ball?)

Short Answer

Expert verified
The baseball's speed is 154.44 km/h.

Step by step solution

01

Understanding the Doppler Effect

We need to consider that the radar measures a Doppler shift in frequency upon detection, meaning the frequency changes due to the motion of the baseball. This shift occurs twice: first when the radar waves hit the baseball and again when the waves reflect back to the detector.
02

Determine the Formula for Speed Calculation

The formula for Doppler shift in frequency due to relative motion is given by:\[ \Delta f = \frac{2v}{c} f_0 \]where \(v\) is the speed of the baseball, \(c\) is the speed of light (approximately \(3 \times 10^8\) m/s), and \(\Delta f / f_{0} = 2.86 \times 10^{-7}\) is the given fractional frequency shift. The factor of 2 accounts for the double shift.
03

Rearrange the Formula to Solve for the Baseball's Speed

Rearranging the formula \( \Delta f = \frac{2v}{c} f_0 \) to solve for \(v\):\[ v = \frac{\Delta f}{f_{0}} \times \frac{c}{2} \]
04

Calculate the Baseball's Speed

Substitute the given values:\[ v = 2.86 \times 10^{-7} \times \frac{3 \times 10^{8}}{2} \]Simplifying this gives:\[ v = 2.86 \times 10^{-7} \times 1.5 \times 10^{8} \]\[ v = 4.29 \times 10^{1}\]\[ v = 42.9 \text{ m/s} \]
05

Convert Speed to Kilometers Per Hour

To convert the speed from meters per second to kilometers per hour, use the conversion:\[ 1 \text{ m/s} = 3.6 \text{ km/h} \]Therefore, \(v = 42.9 \text{ m/s} \times 3.6 = 154.44 \text{ km/h} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Waves
Electromagnetic waves are vital tools for various technologies in our everyday lives. They are waves of electric and magnetic fields that travel through space at the speed of light, approximately \(3 \times 10^8\) meters per second.
Unlike sound or water waves, electromagnetic waves do not need a medium to travel, making them essential for communication, such as in radar systems.
A radar device specifically uses radio waves, which are a type of electromagnetic wave. These waves are emitted, bounce off a moving object such as a baseball, and return as echoes. The change in the frequency of these returning waves can reveal the speed of the object. This is because the motion of the object affects the waves, a phenomenon described by the Doppler Effect.
Frequency Shift
The Doppler Effect is responsible for the frequency shift of electromagnetic waves. When an object moves toward or away from an observer, the frequency of the waves it reflects changes. This change is known as the frequency shift.
For the Doppler Effect, there are two primary shifts to consider. First, when waves emitted by the radar hit the moving baseball, and then when waves are reflected back. This reflects a double shift rather than just a single alteration.
  • The formula for the frequency shift is: \[ \Delta f = \frac{2v}{c} f_0 \]
  • \(\Delta f\) is the difference in frequency.
  • \(v\) is the speed of the baseball.
  • \(c\) is the speed of light.
  • \(f_0\) is the original frequency emitted by the radar.

Given the fractional shift \(\Delta f / f_{0} = 2.86 \times 10^{-7}\), one can rearrange the formula to find the speed of the baseball.
Speed Calculation
Once the relationship between the frequency shift and velocity is established, you can calculate the speed of an approaching baseball. The aim is to find \(v\), which represents the speed of the baseball. Using the formula:
\[ v = \frac{\Delta f}{f_{0}} \times \frac{c}{2} \]
Given the fractional frequency shift \(\Delta f / f_{0} = 2.86 \times 10^{-7}\), the Doppler shift formula can be directly applied. Substituting the known values:\
\[ v = 2.86 \times 10^{-7} \times \frac{3 \times 10^{8}}{2} \]
Yields
\[ v = 42.9 \text{ m/s} \]
Converting from meters per second to kilometers per hour involves multiplying by 3.6 (this converts velocity to a more common unit for speed, as \(1 \text{ m/s} = 3.6 \text{ km/h}\)):
\[ v = 42.9 \text{ m/s} \times 3.6 = 154.44 \text{ km/h} \]
Therefore, the baseball is traveling at approximately 154.44 km/h.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

An enemy spaceship is moving toward your starfighter with a speed, as measured in your frame, of 0.400 \(\mathrm{c}\) . The enemy ship fires a missile toward you at a speed of 0.700\(c\) relative to the enemy ship (Fig. 37.28\()\) . (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure that the enemy ship is \(8.00 \times 10^{6} \mathrm{km}\) away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?

A \(0.100-\mu g\) speck of dust is accelerated from rest to a speed of 0.900\(c\) by a constant \(1.00 \times 10^{6} \mathrm{N}\) force. (a) If the nonrelativistic form of Newton's second law \((\Sigma F=m a)\) is used, how far does-the object travel to reach its final speed? (b) Using the correct relativistic treatment of Section 37.8 , how far does the object travel to reach its final speed? (c) Which distance is greater? Why?

One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is \(\lambda=656.3 \mathrm{nm},\) in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to \(\lambda=953.4 \mathrm{nm}\) , in the infrared portion of the spectrum. How fast are the emitting atoms moving relative to the earth? Are they approaching the earth or receding from it?

An observer in frame \(S^{\prime}\) is moving to the right \((+x-\text { direction })\) at speed \(u=0.600 \mathrm{c}\) away from a stationary observer in frame \(S\) . The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in S measure for the particle if \((a) v^{\prime}=0.400 c ;(b) v^{\prime}=0.900 c\) (c) \(v^{\prime}=0.990 c ?\)
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Force

Read Explanation

Dynamics

Read Explanation

Engineering Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.